Question

Find the function$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$of the form $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{a}{\mathit{e}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathit{b}{\mathit{e}}^{\mathbf{2}\mathbf{t}}$such that $\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$and$\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$.

Short answer

The function of the form $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$such that $f\left(0\right)=1$and $f\text{'}\left(0\right)=4$is $f\left(t\right)=2e^{3t}-e^{2t}.$

Step-by-step solution

Consider the points and substitute these in the standard equation

Consider the function $f\left(t\right)=a{e}^{3t}+b{e}^{2t}$. Put$f\left(0\right)=1$ in$f\left(t\right)=a{e}^{3t}+b{e}^{2t}$

$$\begin{gathered} f\left(0\right)=a{e}^{3\left(0\right)}+b{e}^{2\left(0\right)} \\ 1=a+b \end{gathered}$$

Consider the derivative of the polynomial

Take the derivative of the polynomial$f\left(t\right)=a{e}^{3t}+b{e}^{2t}$ and put$f\text{'}\left(0\right)=4$ into its derivative.

$$\begin{gathered} f^{\text{'}}=\left(t\right)=3a{e}^{3t}+2b{e}^{2t} \\ {f}^{\text{'}}\left(0\right)=3a{e}^{3\times 0}+2b{e}^{2\times 0} \\ 4=3a+2b \end{gathered}$$

Solve the above equations.

Arrange the equations obtained in above steps into the matrix form.

$\left(\begin{array}{cc}1& 1\\ 3& 2\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}1\\ 4\end{array}\right)$

Upon solving the values of$a,b$ are obtained as$a=2,b=-1$

Substitute these values in the polynomial equation.

role="math" localid="1659340662448" $$\begin{gathered} f\left(t\right)=a{e}^{3t}+b{e}^{2t} \\ f\left(t\right)=2e^{3t}-1e^{2t} \end{gathered}$$

The function$f\left(t\right)$ of the form$f\left(t\right)=a{e}^{3t}+b{e}^{2t}$ such that$f\left(0\right)=1$ and$f\text{'}\left(0\right)=4$ is $f\left(t\right)=2e^{3t}-e^{2t}$.